1
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$, $\vec{a} \cdot \vec{b} < 0$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then the value of $\sin\theta + \tan\theta$ is...
A
$\dfrac{\sqrt{2} - 2}{2}$
B
$\dfrac{\sqrt{2} + 2}{2}$
C
$\dfrac{1 + \sqrt{2}}{2}$
D
$\dfrac{2 - \sqrt{2}}{2}$
2
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
Let $\vec{a} = \lambda\hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}, \vec{c} = \hat{i} + \mu\hat{j} + \hat{k}$
If $\vec{a}$ is parallel to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$ then $\lambda - \mu = \ldots$
A
$-2$
B
$-1$
C
$2$
D
$1$
3
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The equation of the perpendicular line from the point $(2,-3,1)$ to the line $\dfrac{x + 1}{2} = \dfrac{y - 3}{3} = \dfrac{z + 2}{-1}$ is
A
$\dfrac{x - 2}{24} = \dfrac{y + 3}{13} = \dfrac{z - 1}{9}$
B
$\dfrac{x - 2}{24} = \dfrac{y - 3}{-13} = \dfrac{z - 1}{9}$
C
$\dfrac{x + 2}{-24} = \dfrac{y + 3}{13} = \dfrac{z + 1}{-9}$
D
$\dfrac{x - 2}{-24} = \dfrac{y + 3}{13} = \dfrac{z - 1}{-9}$
4
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The angle between the line $x - 1 = 2 - y = \dfrac{2z - 6}{4}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is
A
$\dfrac{\pi}{3}$
B
$\dfrac{\pi}{6}$
C
$\dfrac{\pi}{4}$
D
$\dfrac{\pi}{12}$

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